# An object's two dimensional velocity is given by v(t) = ( sin(pi/3t) , 2cos(pi/2t )- t ). What is the object's rate and direction of acceleration at t=5 ?

Jan 4, 2018

The rate of acceleration is $= 4.17 m {s}^{-} 2$ in the direction ${97.2}^{\circ}$ anticlockwise from the $\text{x-axis}$

#### Explanation:

The acceleration is the derivative of the velocity

$v \left(t\right) = \left(\sin \left(\frac{\pi}{3} t\right) , 2 \cos \left(\frac{\pi}{2} t\right) - t\right)$

The acceleration is

$a \left(t\right) = v ' \left(t\right) = \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3} t\right) , - \pi \sin \left(\frac{\pi}{2} t\right) - 1\right)$

When $t = 5$

$a \left(5\right) = v ' \left(5\right) = \left(\frac{\pi}{3} \cos \left(\frac{\pi}{3} \cdot 5\right) , - \pi \sin \left(\frac{\pi}{2} \cdot 5\right) - 1\right)$

$= \left(0.52 , - 4.14\right)$

The rate of acceleration is

$| | a \left(5\right) | | = \sqrt{{\left(0.52\right)}^{2} + {\left(- 4.14\right)}^{2}} = 4.17 m {s}^{-} 2$

And the direction is

$\theta = \arctan \left(- \frac{4.14}{0.52}\right) = {97.2}^{\circ}$ anticlockwise from the $\text{x-axis}$